Harvard-MIT November Tournament

Solution:

Let $\theta = \angle BAC$, and $\cos \theta = \frac{7}{9}$ implies $\cos \frac{\theta}{2} = \sqrt{\frac{1 + \frac{7}{9}}{2}} = \frac{2\sqrt{2}}{3}$; $\sin \frac{\theta}{2} = \frac{1}{3}$; $BC = \sqrt{4 + 25 - 2(2)(5) \frac{7}{9}} = \frac{11}{3}$.

Let $O_1$ be the excircle of $\triangle ABC$ tangent to lines $AB$ and $AC$, and let $r_1$ be its radius; let $O_1$ be tangent to line $AB$ at point $P_1$. Then $AP_1 = \frac{AB + BC + CA}{2}$ and $\frac{r_1}{AP_1} = \tan \frac{\theta}{2} = \frac{1}{2\sqrt{2}} \Longrightarrow r_1 = \frac{16}{3 \cdot 2\sqrt{2}}$.

Let $O_n$ be a circle tangent to $O_{n-1}$ and the lines $AB$ and $AC$, and let $r_n$ be its radius; let $O_n$ be tangent to line $AB$ at point $P_n$. Then $\frac{O_nP_n}{AO_n} = \sin \frac{\theta}{2} = \frac{1}{3}$; since $\triangle AP_nO_n \sim \triangle AP_{n-1}O_{n-1}$ and $O_nO_{n-1} = r_n + r_{n-1}$, we have

$$\frac{1}{3} = \frac{O_nP_n}{AO_n} = \frac{r_n}{AO_{n-1} + O_{n-1}O_n} = \frac{r_n}{3r_{n-1} + r_n + r_{n-1}} \Longrightarrow r_n = 2r_{n-1} = 2^{n-1} \frac{16}{3 \cdot 2\sqrt{2}}.$$

We want the highest $n$ such that $O_n$ is contained inside $\triangle ADE$. Let the incircle of $\triangle ADE$ be tangent to $AD$ at $X$; then the inradius of $\triangle ADE$ is
$$\frac{AX}{\tan \frac{\theta}{2}} = \frac{\frac{500 + 200 - \frac{1100}{2}}{2\sqrt{2}}}{2} = \frac{500}{3 \cdot 2\sqrt{2}}.$$

We want the highest $n$ such that $r_n \leq \frac{500}{3 \cdot 2\sqrt{2}}$; thus $2^{n-1} \cdot 16 \leq 500 \Longrightarrow n = 5$.